Question

An electron will have highest energy in which set ? A. 3,2,1,1/2 B. 4,2,-1,1/2 C. 4,1,0,-1/2 D. 5,0,0,1/2

Answer 1

B. 4, 2, -1, `1/2`

Explanation:

`(n + l) ∝ "Energy"`

Greater the `(n + l)` value greater is the energy of electron.

If two electrons have same `(n + l)` value then the one with greater `n` value is of more energy.

A. `(n + l) = (3 + 2) = 5`

B. `(n + l) = (4 + 2) = 6`

C. `(n + l) = (4 + 1) = 5`

D. `(n + l) = (5 + 0) = 5`

Therefore, B is the correct option.

Answer 2

We must identify the orbital first...

Then we would find that the `5s` are higher in energy than the `4d` MOST of the time. The only exception I know is yttrium, which has the `4d` orbitals SLIGHTLY higher than the `5s` (by `"0.21 eV"`, or `"4.84 kcal/mol"`).

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More info on quantum numbers is already given here.

`A)` `(n,l,m_l,m_s) = (3,2,1,1/2)`

This indicates a `3d` orbital of some sort, with a spin-up electron.

`B)` `(n,l,m_l,m_s) = (4,2,-1,1/2)`

This indicates a `4d` orbital of some sort, with a spin-up electron.

`C)` `(n,l,m_l,m_s) = (4,1,0,-1/2)`

This indicates a `4p_z` orbital, with a spin-down electron.

`D)` `(n,l,m_l,m_s) = (5,0,0,1/2)`

This indicates a `5s` orbital, with a spin-up electron.

Data from Inorganic Chemistry (Miessler, Fischer, and Tarr) shows that the `5s` are clearly higher in energy than the `4d` for MOST transition metals that have valence electrons in those orbitals.

Only yttrium apparently has `4d` orbitals SLIGHTLY higher in energy than the `5s`. The rest of the transition metals, zirconium through cadmium, have the `5s` orbital higher in energy than the `4d`.